Problem: Let $a$ and $b$ be real numbers such that
\[a^3 - 15a^2 + 20a - 50 = 0 \quad \text{and} \quad 8b^3 - 60b^2 - 290b + 2575 = 0.\]Compute $a + b.$
Let $x = a - 5.$  Then $a = x + 5,$ so
\[(x + 5)^3 - 15(x + 5)^2 + 20(x + 5) - 50 = 0,\]which simplifies to $x^3 - 55x - 200 = 0.$

Let $y = b - \frac{5}{2}.$  Then $b = y + \frac{5}{2},$ so
\[8 \left( y + \frac{5}{2} \right)^3 - 60 \left( y + \frac{5}{2} \right)^2 - 290 \left( y + \frac{5}{2} \right) + 2575 = 0,\]which simplifies to $y^3 - 55y + 200 = 0.$  (Note that through these substitutions, we made the quadratic term vanish in each of these cubic equations.)

Consider the function $f(t) = t^3 - 55t.$  Observe that the polynomial $f(t)$ has three roots 0, $\sqrt{55},$ and $-\sqrt{55}.$  Its graph is shown below.

[asy]
unitsize (0.2 cm);

real cubic (real x) {
  return ((x^3 - 55*x)/12);
}

draw(graph(cubic,-8.5,8.5));
draw((-18,0)--(18,0));
draw((0,-18)--(0,18));

dot("$\sqrt{55}$", (sqrt(55),0), SE);
dot("$-\sqrt{55}$", (-sqrt(55),0), SW);
[/asy]

Let $0 \le t \le \sqrt{55}.$  Then
\[[f(t)]^2 = (t^3 - 55t)^2 = t^2 (t^2 - 55)^2 = t^2 (55 - t^2)^2 = t^2 (55 - t^2)(55 - t^2).\]By AM-GM,
\[2t^2 (55 - t^2)(55 - t^2) \le \left( \frac{(2t^2) + (55 - t^2) + (55 - t^2)}{3} \right)^3 = \left( \frac{110}{3} \right)^3 < 40^3,\]so
\[[f(t)]^2 < 32000 < 32400,\]which means $|f(t)| < 180.$

Since $f(t)$ is an odd function, $|f(t)| < 180$ for $-\sqrt{55} \le t \le 0$ as well.  This means that the equation $f(t) = 200$ has exactly one real root.  Similarly, $f(t) = -200$ has exactly one real root.  Furthermore, since $f(t)$ is an odd function, these roots add up to 0.

Then
\[a - 5 + b - \frac{5}{2} = 0,\]so $a + b = 5 + \frac{5}{2} = \boxed{\frac{15}{2}}.$